Problem: Positive integers $a$, $b$, and $c$ are chosen so that $a<b<c$, and the system of equations \[
2x + y = 2003\quad\text{and}\quad y = |x-a| + |x-b| + |x-c|
\]has exactly one solution. What is the minimum value of $c$?
Solution: Since the system has exactly one solution, the graphs of the two equations must intersect at exactly one point. If $x<a$, the equation $y = |x-a| + |x-b| + |x-c|$ is equivalent to $y =-3x + (a+b+c)$. By similar calculations we obtain

\[
y =
\begin{cases}
-3x + (a+b+c), &\text{if }x<a\\
-x + (-a+b+c), &\text{if }a\le x<b\\
x + (-a-b+c), &\text{if }b\le x<c\\
3x + (-a-b-c), &\text{if }c\le x.
\end{cases}
\]Thus the graph consists of four lines with slopes $-3$, $-1$, 1, and 3, and it has corners at $(a, b+c-2a)$, $(b, c-a)$, and $(c,
2c-a-b)$.

On the other hand, the graph of $2x+y = 2003$ is a line whose slope is $-2$. If the graphs intersect at exactly one point, that point must be $(a, b+c-2a).$ Therefore

$ 2003 = 2a + (b+c-2a) = b+c. $

Since $b<c$, the minimum value of $c$ is $\boxed{1002}$.